On Laplace transform of periodic functions

Question

I recently bumped into this theorem regarding the Laplace transform of periodic function:

Given a periodic function $f(t)=f(t+p)$, where $p$ is its period, then its Laplace transform is given by:

$$\dfrac{1}{1-e^{-ps}}\int_0^p f(t)e^{-st}dt $$

I found it rather strange looking, so I decided to prove it myself from the definition, that is to prove:

$$\dfrac{1}{1-e^{-ps}}\int_0^p f(t)e^{-st}dt =\int_0^\infty f(t)e^{-st}dt $$

First I will evaluate the LHS:

Consider the indefinite integral $$\int f(t)e^{-st}dt$$ By using integration by parts, this integral will have the following form:

$$e^{-st}g_0(s)I_0(t)+e^{-st}g_1(s)I_1(t)+...+e^{-st}g_n(s)I_n(t)$$

Now since $$\int_0^\infty f(t)e^{-st}dt $$ can be rewritten as $$\int_0^p f(t)e^{-st}dt + \int_p^{2p} f(t)e^{-st}dt +\int_{2p}^{3p} f(t)e^{-st}dt+... $$

Therefore for the first term $e^{-st}g_0(s)I_0(t)$ we have

$$e^{-sp}g_0(s)I_0(p)-g_0(s)I_0(0)+e^{-2ps}g_0(s)I_0(2p)-e^{-ps}g_0(s)I_0(p)+...=-g_0(s)I_0(0)$$ since all terms cancel out except for $-g_0(s)I_0(0)$

The same goes for the other terms $e^{-st}g_1(s)I_1(t),e^{-st}g_2(s)I_2(t)...,e^{-st}g_n(s)I_n(t)$

Therefore $$\int_0^\infty f(t)e^{-st}dt= -g_0(s)I_0(0)-g_1(s)I_1(0)-...-g_n(s)I_n(0) $$

This is the first half of proof.

Now I come to the RHS of the equation: $$\dfrac{1}{1-e^{-ps}}\int_0^p f(t)e^{-st}dt $$

$$\dfrac{1}{1-e^{-ps}}$$ is a geometric series with common ratio $e^{-ps}$ therefore the RHS can be rewritten as :

$$(1+e^{-ps}+e^{-2ps}+e^{-3ps}+...)\int_0^p f(t)e^{-st}dt$$

And since $$\int_0^p f(t)e^{-st}dt=[e^{-ps}g_0(s)I_0(p)-g_0(s)I_0(0)]+[e^{-ps}g_1(s)I_1(p)-g_1(s)I_1(0)]+...+[e^{-ps}g_n(s)I_n(p)-g_n(s)I_n(0)]$$

Therefore the RHS is $$(1+e^{-ps}+e^{-2ps}+e^{-3ps}+...)(e^{-ps}g_0(s)I_0(p)-g_0(s)I_0(0)+e^{-ps}g_1(s)I_1(p)-g_1(s)I_1(0)+...+e^{-ps}g_n(s)I_n(p)-g_n(s)I_n(0) )$$

By distributing the two parentheses we'll have:

$$ g_0(s)I_0(p)(e^{-ps}+e^{-2ps}+e^{-3ps}+...)- g_0(s)I_0(0)(1+e^{-ps}+e^{-2ps}+...)+g_1(s)I_1(p)(e^{-ps}+e^{-2ps}+e^{-3ps}+...)- g_1(s)I_1(0)(1+e^{-ps}+e^{-2ps}+...)+...+g_n(s)I_n(p)(e^{-ps}+e^{-2ps}+e^{-3ps}+...)- g_n(s)I_n(0)(1+e^{-ps}+e^{-2ps}+...) $$

Now for the above expression to evaluate to the LHS $$g_0(s)I_0(0)-g_1(s)I_1(0)-...-g_n(s)I_n(0) $$

This will only occur when :

$I_0(0)=I_0(p), I_1(0)=I_1(p),...,I_n(0)=I_n(p) $

That is for all $I(t)$, it has to be the case that $I(0)=I(p)$

So that $$\dfrac{1}{1-e^{-ps}}\int_0^p f(t)e^{-st}dt =\int_0^\infty f(t)e^{-st}dt $$ is only valid when $I(0)=I(p)$

Therefore I concluded that this theorem is not applicable for all periodic functions:

For example if we have: $f(t)=t$ for all $0=<t<=1 $ and define $f(t)=f(t+1)$

Then we cannot apply the theorem since $$\int f(t)e^{-st}dt=\dfrac{-te^{-st}}{s}-\dfrac{e^{-st}}{s^2}$$

And it's obvious that for $I_0(t)=t$, $I_0(0)$ is not equal to $I_0(1)$

Therefore the theorem is not applicable in this case.

So is my analysis correct? is this theorem not applicable for all periodic functions?

Answer 1

Let $f\colon[0,\infty)\to\mathbb{R}$ b periodic of period $p$. $$\begin{align} \int_0^\infty f(t)\,e^{-st}\,dt&=\sum_{k=0}^\infty\int_{pk}^{p(k+1)} f(t)\,e^{-st}\,dt\\ &=\sum_{k=0}^\infty\int_{0}^{p} f(t+k\,p)\,e^{-s(t+kp)}\,dt\\ &=\Bigl(\sum_{k=0}^\infty e^{-skp}\Bigr)\int_{0}^{p} f(t)\,e^{-st}\,dt\\ &=\frac{1}{1-e^{-sp}}\int_{0}^{p} f(t)\,e^{-st}\,dt. \end{align}$$ The formula is valid for all periodic functions.

Answer 2

Here is another way:

Suppose $f$ has period $T$. Then,

$$F(s)=\int_{0}^{T}f(t)e^{-st}dt+\int_{T}^{\infty }f(t)e^{-st}dt=\int_{0}^{T}f(t)e^{-st}dt+\int_{T}^{\infty }f(t-T)e^{-st}dt$$.

Let $\tau =t-T$ so that $$F(s)=\int_{0}^{T}f(t)e^{-st}dt+\int_{0}^{\infty }f(\tau)e^{-s(\tau +T)}d\tau=\int_{0}^{T}f(t)e^{-st}dt+e^{-sT}\int_{0}^{\infty }f(\tau)e^{-s\tau }d\tau =\int_{0}^{T}f(t)e^{-st}dt+e^{-sT}F(s)$$ which, upon solving for $F(s)$ gives the result:

$F(s)= \frac{1}{1-e^{-sT}}\int_{0}^{T} f(t)\,e^{-st}\,dt$